Step 1. Now, $A, X$ are path connected so $\tilde{H}_0(A) =\tilde{H}_0(X) = 0$. So $\tilde{H}_0(X,A)=0$ by the long exact sequence, which gives us that $H_0(X,A) \cong \mathbb{Z}$.

Step 2. The map $i_*: H_1(A) \to H_1(X)$ takes $[\alpha] \mapsto [i(\alpha)]$ where $i$ is the inclusion. In other words, it is injective, and so $0 = \ker i_* = \text{im}\:f$, and so $f$ is the zero map. This then leaves two short exact sequences, and implies that $H_2(X,A) \cong H_2(X) \cong \mathbb{Z}$.

Step 3. It is an immediate consequence of the other remaining short exact sequence that $H_1(X,A) \cong H_1(X)/\text{im }i_* \cong \mathbb{Z} \oplus \mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}$.

What I'm looking for are mainly comments on if this was done properly, but I also have two specific queries:

1. Have I used the reduced homology 'version' of the long exact sequence correctly? I'm very iffy with reduced homology...

2. My reason for $i_*$ being injective was a little handwavy. Is it always the case that a homomorphism induced by inclusion of spaces $A\hookrightarrow X$ is injective?

2 Answers
2

Overall, this looks pretty good. You probably want to point out that $H_2(A)\cong0$.

Specifically:

Yes. You wrote the sequence down properly. Reduced homology is largely used to make computations easier. It arises from augmenting the (singular, simplicial, etc.) chain complex associated to a space with an additional map $\cdots\to C_0(X)\xrightarrow{\epsilon}\mathbb{Z}\to 0$ where $\epsilon(\sum_i n_i\sigma_i)=\sum_i n_i$.

It is not always the case that the map induced by an inclusion of spaces is an injection on homology. Consider the inclusion $S^1\hookrightarrow D^2$. The disc is contractible, and so has the same homology as a point, where as $S^1$ has nontrivial $H_1$. In your case, you can see that the inclusion of $S^1$ into $S^1\times S^1$ as $S^1\times \{*\}$ induces an inclusion on homology by giving both spaces suitable cell-complex structures (be it CW-, $\Delta$-, or simplicial-) and explicitly following the map induced on the chain complex level to homology.

I can't comment because I don't have enough points, but I'd like to remark that $\tilde H_n(X,A)$ is the same as $H_n(X,A)$ for all $n$ when $A\ne \emptyset$, so this step 1 conclusion seems incorrect...